# What do elements of CKM matrix imply?

In CKM matrix, there are 9 elements, e.g. Vud = .974, Vus = .227 ,Vub = .004. The sum of these 3 elements is greater than one, so they cannot represent the probability of an up quark to transform in an interaction/decay into down, strange, bottom quark respectively by emitting W+ boson. Then what do these elements imply?

https://en.wikipedia.org/wiki/Cabibbo%E2%80%93Kobayashi%E2%80%93Maskawa_matrix $$\begin{bmatrix} d^\prime \\ s^\prime \\ b^\prime \end{bmatrix} = \begin{bmatrix} V_{ud} & V_{us} & V_{ub} \\ V_{cd} & V_{cs} & V_{cb} \\ V_{td} & V_{ts} & V_{tb} \end{bmatrix} \begin{bmatrix} d \\ s \\ b \end{bmatrix}$$ The d,s,b quarks are eigenstates of mass. The d',s',b' states are states that W,Z leave within their own SU(2) doublets. For example, $W^+$ takes d' into u and nothing else. So, the CKM matrix V says d' is a linear combination of d, s, and b. The coefficients are "amplitudes" not probabilities. The probabilities are ${|amplitude|}^2$. V is unitary ($V V^{\dagger} =I$) to conserve probability. Therefore, $$|V_{ud}|^2+|V_{us}|^2+|V_{ub}|^2=1$$ $$.974^2+.224^2+.004^2=.999$$ This is 1 within experimental error and can be used to argue that another generation of quarks is not needed. You can see all the experimentally measured CKM values with their errors at the above link. By calculating $V V^{\dagger}$ you can check for yourself how well conservation of probability is satisfied.